In Fig. 16.201, ABCD is a quadrilateral inscribed in a circle with centre O. CD is produced to E such that AED = 95° and OBA = 30°. Find OAC.

ADE = 95o (Given)

Since,


OA = OB, so


OAB = OBA


OAB = 30o


ADE + ADC = 180o (Linear pair)


95o + ADC = 180o


ADC = 85o


We know that,


ADC = 2 ADC


ADC = 2 * 85o


ADC = 170o


Since,


AO = OC (Radii of circle)


OAC = OCA (Sides opposite to equal angle) (i)


In triangle OAC,


OAC + OCA + AOC = 180o


OAC + OAC + 170o = 180o [From (i)]


2 OAC = 10o


OAC = 5o


Thus,


OAC = 5o


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